Estimate Pi Using Conductive Paint

Given an "infinite" 2D square grid of resistors with resistance R, the equivalent resistance across the diagonal of one square is 2R/pi [1,2]. Working out this solution theoretically has sniped many a nerd. Furthermore, it is prohibitively laborious for an amateur to attempt to verify this experimentally -- until now. Using Bare Paint, I drew a square grid of resistors on a page of normal copy paper, and measured resistances with a multimeter. My wishful thinking was that, with a large enough array of resistors (14 x 14), I could approximate "infinity" sufficiently from the perspective of squares in the middle of the page. Drawing the resistors with the Bare Paint seemed easier than gathering and connecting 196 resistors.

While one set of measurements yielded a value of pi as impressive (to me) as 3.38, overall I observed poor consistency of measurements among adjacent squares near the center of the page, and poor repeatability of measurements for the squares tested. Nevertheless, I hope that you, dear reader, will consider trying this experiment for yourself. I will tell you how I did it, and perhaps you will take more care than I to apply the paint accurately and precisely; otherwise, you will too observe a high variance of resistance among your resistors. Note also that the resistance of the Bare Paint decreases as it dries, so be sure to allow ample time for drying (tens of minutes) before measurements.

--
Submitted by Ace Monster Toys Hackerspace in Oakland, CA for the Instructables Sponsorship Program

I used one tube of Bare Paint, a multimeter with leads (pictured in the intro section), and an 8 1/2" x 11" sheet of paper.

To make a square grid of lines to guide your eyes and the pen when drawing the resistors, you can first fold the sheet diagonally to be flush with a long edge, as shown. Cut/tear the sheet along the interface between the two-layer part and the still-one-layer part to end up with (after unfolding) a square piece (8 1/2" x 8 1/2") of paper.

Then, for each of the two axes of the sheet, successively fold the sheet in half four times and unfold it completely.

With a calm and steady hand, keeping an eye on the guide lines, and with light pressure on the pen body, draw the grid. Well, to decrease variability, you may try a different method, but drawing by hand is easy. I waited a bit after drawing all lines in one direction before drawing the crossing lines, but I think that was unwise because the drawn lines were able to dry/harden a bit and thus more care was required on my part when drawing across the already-drawn lines -- sometimes the pen seemed to "jump" with some momentum over the already-drawn lines, and thus I had to backtrack to ensure well-connected junctions.

You can clearly see than my drawn grid lacked uniformity, but I didn't know then how much that would matter. Hopefully you will do better than me.

Also, note how I drew "dangling" resistors at the edges of the sheet. These are useful to estimate the resistance of an isolated stretch of the paint equal in length to a side of one square; you need this value to estimate pi.

"To measure is to know." -Lord Kelvin

There are three simple measurements to do. Using your multimeter in "ohmmeter" mode (the units of resistance are Ohms, symbolized by the Greek capital letter Omega), you can measure (1) the resistance across an isolated "dangling" resistor that is not part of the grid; let's call this resistance R. Next, you can measure (2) the resistance across one side of a square near the center of the paper; the resistance should theoretically be R_side = R / 2. Finally, you can measure (3) the resistance across the diagonal of one square; the resistance should theoretically be R_diag = R * (2 / pi). So, pi should equal 2 * (R / R_diag). Is it close to 3.14? Note that I can list at most three significant digits because my pictured multimeter only provided three. I got as close as 3.38 for one set of measurements.